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Question Number 134526 by mohammad17 last updated on 04/Mar/21

Answered by mr W last updated on 05/Mar/21

$$\frac{1}{1+x}=1-x+x^2-x^3+...$$$$\int \frac{1}{1+x}dx=\int (1-x+x^2-x^3+...)dx$$$$=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...+C$$$$=\ln (1+x)+C$$

Commented by mathmax by abdo last updated on 05/Mar/21

$$\mathrm{sir}\mathrm{mrw}\mathrm{the}\mathrm{formula}\frac{1}{1+\mathrm{x}}=1-\mathrm{x}+{\mathrm{x}}^2-....\mathrm{is}\mathrm{valable}\mathrm{for}\mid \mathrm{x}\mid <1...!$$

Commented by mr W last updated on 05/Mar/21

$$yes,iknow.basicallyyouare$$$$absolutelyright.butbothfor\mid x\mid <1$$$$andfor\mid x\mid >1wegetthesame.sowe$$$$canheretreatxjustasasymbol$$$$despiteitsconcretevalues.$$$$heretheprooffor\mid x\mid >1:$$$$\mid x\mid >1\Rightarrow \mid \frac{1}{x}\mid <1$$$$\frac{1}{1+x}=\frac{\frac{1}{x}}{1+\frac{1}{x}}=1-\frac{1}{1+\frac{1}{x}}=1-(1-\frac{1}{x}+\frac{1}{x^2}-\frac{1}{x^3}+...)$$$$\frac{1}{1+x}=\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+...$$$$\int \frac{1}{1+x}dx=\ln x+(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+...)+C$$$$\int \frac{1}{1+x}dx=\ln x+\ln (1+\frac{1}{x})+C$$$$\int \frac{1}{1+x}dx=\ln (1+x)+C(\mid x\mid >1)$$

Commented by mathmax by abdo last updated on 06/Mar/21

$$\mathrm{yes}$$

Answered by mathmax by abdo last updated on 05/Mar/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\int }_{\mathrm{x}}^1\frac{\mathrm{dt}}{1+\mathrm{t}}\mathrm{if}\mathrm{o}<\mathrm{x}<1\Rightarrow \mathrm{o}⩽\mathrm{t}⩽1\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\int }_{\mathrm{x}}^1\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_{\mathrm{x}}^1{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}{\left[{\mathrm{t}}^{\mathrm{n}+1}\right]}_{\mathrm{x}}^1$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{1-{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}$$$$=-\ln (1+\mathrm{x})+\ln \left(2\right)$$$$\Rightarrow {\int }_1^{\mathrm{x}}\frac{\mathrm{dt}}{1+\mathrm{t}}=\ln (1+\mathrm{x})-\ln \left(2\right)(\mathrm{c}=-\ln 2)$$$$\mathrm{if}\mathrm{x}>1\mathrm{we}\mathrm{take}\mathrm{g}\left(\mathrm{x}\right)={\int }_1^{\mathrm{x}}\frac{\mathrm{dt}}{1+\mathrm{t}}{=}_{\mathrm{t}=\frac{1}{\mathrm{u}}}{\int }_1^{\frac{1}{\mathrm{x}}}\frac{-\mathrm{du}}{{\mathrm{u}}^2(1+\frac{1}{\mathrm{u}})}$$$$=-{\int }_1^{\frac{1}{\mathrm{x}}}\frac{\mathrm{du}}{{\mathrm{u}}^2+\mathrm{u}}={\int }_{\frac{1}{\mathrm{x}}}^1\frac{\mathrm{du}}{\mathrm{u}(\mathrm{u}+1)}={\int }_{\frac{1}{\mathrm{x}}}^1\frac{1}{\mathrm{u}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\mathrm{du}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_{\frac{1}{\mathrm{x}}}^1{\mathrm{u}}^{\mathrm{n}-1}\mathrm{du}=$$$$={\int }_{\frac{1}{\mathrm{x}}}^1\frac{\mathrm{du}}{\mathrm{u}}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\left[\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\right]}_{\frac{1}{\mathrm{x}}}^1$$$$=\ln \left(\mathrm{x}\right)+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}(1-\frac{1}{{\mathrm{x}}^{\mathrm{n}}})$$$$=\ln \left(\mathrm{x}\right)+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}}$$$$=\ln \left(\mathrm{x}\right)-\ln \left(2\right)+\ln (1+\frac{1}{\mathrm{x}})$$$$=\ln (1+\mathrm{x})-\ln \left(2\right)$$

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